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Center of mass of continuous objects

If the object in question has its mass distributed uniformly in space, rather than as a collection of discrete particles, then m j d m , and the summation becomes an integral:

r CM = 1 M r d m .

In this context, r is a characteristic dimension of the object (the radius of a sphere, the length of a long rod). To generate an integrand that can actually be calculated, you need to express the differential mass element dm as a function of the mass density of the continuous object, and the dimension r . An example will clarify this.

Cm of a uniform thin hoop

Find the center of mass of a uniform thin hoop (or ring) of mass M and radius r .

Strategy

First, the hoop’s symmetry suggests the center of mass should be at its geometric center. If we define our coordinate system such that the origin is located at the center of the hoop, the integral should evaluate to zero.

We replace dm with an expression involving the density of the hoop and the radius of the hoop. We then have an expression we can actually integrate. Since the hoop is described as “thin,” we treat it as a one-dimensional object, neglecting the thickness of the hoop. Therefore, its density is expressed as the number of kilograms of material per meter. Such a density is called a linear mass density    , and is given the symbol λ ; this is the Greek letter “lambda,” which is the equivalent of the English letter “l” (for “linear”).

Since the hoop is described as uniform, this means that the linear mass density λ is constant. Thus, to get our expression for the differential mass element dm , we multiply λ by a differential length of the hoop, substitute, and integrate (with appropriate limits for the definite integral).

Solution

First, define our coordinate system and the relevant variables ( [link] ).

A hoop of radius r is centered on the origin of an x y coordinate system. A short arc of length ds at an angle theta is highlighted and labeled as mass dm. The radius r from the origin to ds is the hypotenuse of the right triangle with bottom side length x.
Finding the center of mass of a uniform hoop. We express the coordinates of a differential piece of the hoop, and then integrate around the hoop.

The center of mass is calculated with [link] :

r CM = 1 M a b r d m .

We have to determine the limits of integration a and b . Expressing r in component form gives us

r CM = 1 M a b [ ( r cos θ ) i ^ + ( r sin θ ) j ^ ] d m .

In the diagram, we highlighted a piece of the hoop that is of differential length ds ; it therefore has a differential mass d m = λ d s . Substituting:

r CM = 1 M a b [ ( r cos θ ) i ^ + ( r sin θ ) j ^ ] λ d s .

However, the arc length ds subtends a differential angle d θ , so we have

d s = r d θ

and thus

r CM = 1 M a b [ ( r cos θ ) i ^ + ( r sin θ ) j ^ ] λ r d θ .

One more step: Since λ is the linear mass density, it is computed by dividing the total mass by the length of the hoop:

λ = M 2 π r

giving us

r CM = 1 M a b [ ( r cos θ ) i ^ + ( r sin θ ) j ^ ] ( M 2 π r ) r d θ = 1 2 π a b [ ( r cos θ ) i ^ + ( r sin θ ) j ^ ] d θ .

Notice that the variable of integration is now the angle θ . This tells us that the limits of integration (around the circular hoop) are θ = 0 to θ = 2 π , so a = 0 and b = 2 π . Also, for convenience, we separate the integral into the x - and y -components of r CM . The final integral expression is

r CM = r CM, x i ^ + r CM, y j ^ = [ 1 2 π 0 2 π ( r cos θ ) d θ ] i ^ + [ 1 2 π 0 2 π ( r sin θ ) d θ ] j ^ = 0 i ^ + 0 j ^ = 0

as expected.

Center of mass and conservation of momentum

How does all this connect to conservation of momentum?

Practice Key Terms 4

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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